Skip to main content
Post Undeleted by S. Carnahan
elaboration.; added 158 characters in body; added 22 characters in body; added 27 characters in body; added 2 characters in body; added 41 characters in body
Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$. I think one possible reason for the apparent incongruity is that lots of complex points of the base-changed space end up more generic after dropping to the original space, because they aren't defined over $\mathbb{Q}$.

Let's consider an explicit example of what happens when we compose a map from $W$ to $W \times X$ along the projection to $X$, when $X = \mathbb{A}^1$. In the base change, we have ideals in $\mathbb{C}[x]$ like $(x-\pi)$ that aren't defined over the integers. These ideals lie in the preimage of $(0) \subset \mathbb{Z}[x]$ under the base change map, since the intersection with the subring is zero.

In higher dimensions, we can have $\mathbb{C}$-points with transcendental coordinates that still satisfy algebraic equations, hence lie on subvarieties, and these will map to the respective generic points under projection.

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$.

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$. I think one possible reason for the apparent incongruity is that lots of complex points of the base-changed space end up more generic after dropping to the original space, because they aren't defined over $\mathbb{Q}$.

Let's consider an explicit example of what happens when we compose a map from $W$ to $W \times X$ along the projection to $X$, when $X = \mathbb{A}^1$. In the base change, we have ideals in $\mathbb{C}[x]$ like $(x-\pi)$ that aren't defined over the integers. These ideals lie in the preimage of $(0) \subset \mathbb{Z}[x]$ under the base change map, since the intersection with the subring is zero.

In higher dimensions, we can have $\mathbb{C}$-points with transcendental coordinates that still satisfy algebraic equations, hence lie on subvarieties, and these will map to the respective generic points under projection.

Post Deleted by S. Carnahan
Source Link
S. Carnahan
  • 45.7k
  • 6
  • 114
  • 220

In general, maps over $W$ from $W$ to $X \times W$ are the same as maps from $W$ to $X$, by the universal property of products. Here, $W = \operatorname{Spec} \mathbb{C}$.